Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Section 15.5 - Triple Integrals in Rectangular Coordinates - Exercises 15.5 - Page 902: 29

Answer

$\frac{16}{3}$

Work Step by Step

$ V=8\int^1_0 \int^{\sqrt{1-x^2}}_0 \int^{\sqrt{1-x^2}}_0 $ dz dy dx =$8\int^1_0 \int^{\sqrt{1-x^2}}_0 \sqrt{1-x^2}$ dy dx =$8\int^1_0 (1-x^2)dx $ =$\frac{16}{3}$
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